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  • When radiation of wavelength  is incident on a metallic surface, the stopping potential is <img alt="4.8" s

When radiation of wavelength \lambda is incident on a metallic surface, the stopping potential is 4.8 volts. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6 volts. Then the threshold wavelength for the surface is -

Option: 1

2\lambda


Option: 2

4\lambda


Option: 3

6\lambda


Option: 4

8\lambda


Answers (1)

best_answer

\begin{aligned} & V_0=\frac{h c}{e \lambda}-\frac{W}{e} \\ & 4.8=\frac{h c}{e \lambda}-\frac{W}{e} \end{aligned}                                           .........................\left ( 1 \right )
1.6=\frac{\mathrm{hc}}{2 \mathrm{e} \lambda}-\frac{\mathrm{W}}{\mathrm{e}}                                         ..........................\left ( 2 \right )
\begin{aligned} & \frac{\mathrm{eq}^{\mathrm{n}}(1)}{\mathrm{eq}^{\mathrm{n}}(2)} \Rightarrow \frac{4.8}{1.6}=\frac{\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\mathrm{W}}{\mathrm{e}}}{\frac{\mathrm{hc}}{2 \mathrm{e} \lambda}-\frac{\mathrm{W}}{\mathrm{e}}} \\ & \frac{3 \mathrm{hc}}{2 \mathrm{e} \lambda}-\frac{3 \mathrm{~W}}{\mathrm{e}}=\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\mathrm{W}}{\mathrm{e}} \\ & \text { or } \frac{2 \mathrm{~W}}{\mathrm{e}}=\frac{\mathrm{hc}}{2 \mathrm{e} \lambda} \\ & \text { or } \frac{2}{\mathrm{e}}\left(\frac{\mathrm{hc}}{\lambda_0}\right)=\frac{\mathrm{hc}}{2 \mathrm{e} \lambda} \\ & \Rightarrow \lambda_0=4 \lambda \end{aligned}

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Pankaj Sanodiya

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